Since $b$ is an upper bound of $S$, then for any $s\in S$, we have $s\le b$, by definition of upper bound.
Take any $t<b$, and assume for the moment (just to see what happens) that there aren't any $s\in S$ with $t<s\le b$. In other words, we're supposing that if $s\in S$, then $s\le t$ or $b<s$. We can't have $b<s$ if $s\in S$ (see first paragraph), so that means we're assuming that $s\le t$ for every $s\in S$. But that means that $t$ is an upper bound of $S$. In particular, then, since $t<b$, then $b$ cannot be the least upper bound of $S$. But $b$ was defined to be the least upper bound of $S$, so our assumption led to an impossible situation (contradiction), and so our assumption was false.
Thus, the desired conclusion holds.